# SOLUTION: log of 9 to the base square root of x - log of 3 to the base x = 9 + 6 log of x to the base square root of 3

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 Question 2570: log of 9 to the base square root of x - log of 3 to the base x = 9 + 6 log of x to the base square root of 3 Answer by kiru_khandelwal(79)   (Show Source): You can put this solution on YOUR website!As, log of a to the base b = log a/log b So, log of 9 to the base square root of x = log 9/log (square root of x) => log 9/log(x^1/2) = > As, log(a^b) = bloga => log 9/log(x^1/2) = log (3^2)/log(x^1/2) = 2log3/((1/2)logx) = 4log3/logx...(1) log of 3 to the base x = log 3/logx...(2) log ofx to the base square root of 3 = log x/log (3^1/2) = log x/ ((1/2)log3) = 2logx/log3...(3) Now according to the quesiton and using the equations (1)(2) and (3) we get 4log3/logx - log3/logx = 9 + 6(2logx/log3) => 3log3/logx = 9 + 12logx/log3 => 3log3/logx = (9log3 + 12logx)/log3 => 3log3*log3 = 9log3*logx + 12logx*logx Let logx be X and log3 be Y => 3Y*Y = 9YX + 12 X*X => 12X^2 + 9XY - 3Y^2 = 0 => 4X^2 + 3XY - Y^2 = 0 => 4X^2 + 4XY-XY -Y^2 = 0 => 4X(X+Y)-Y(X+Y)=0 => (4X-Y)(X+Y)=0 According to the zero product rule 4X-Y =0 and X+Y = 0 => X = Y/4 and X = -Y as Y = log 3 and X = logx So, X= Y/4 => logx = (log 3)/4 = log (3^1/4) => x = 3^1/4 X = -Y => logx = -log 3 = log 3^-1 = log (1/3) so x = 1/3